Giant leap of the space goat This problem was inspired by this Math.SE post  and similar problems involving tethered animals.
A space goat is tethered to the north pole of a spherical asteroid of uniform density and radius $a$ by a rope of length $\pi a$.
Becoming bored with grazing the surface, the goat tests the limits of its freedom by making a series of leaps. What is the longest time that it can remain airborne? The answer should be expressed as a multiple of the orbital period of an object circling the asteroid just above its surface.
Assume that the goat’s strength is not a limiting factor, the asteroid is very much more massive than the goat, atmospheric resistance and the weight of the rope are negligible, and solutions involving any possibility of the rope dragging across the surface are non-optimal.
In the event of the rope suddenly becoming taut assume that it behaves elastically - i.e. no energy is dissipated as heat.
Note: This is an original problem and I don’t yet have a complete solution. It looks doable but don’t necessarily expect a result in closed form.
[Edits: 1. Added assumption about relative masses. 2. Added assumption about behaviour if rope suddenly becomes taut.]
 A: First I assume that any solution which would involve putting tension on the tether would not increase the duration of the jump, since this would only increase the amount of applied force, which would imply that the average angular velocity would be higher and therefore go faster around the asteroid (I have not proven this).
Using this assumption then the solutions would be a Kepler orbit. A second assumption would be that the goat has a negligible mass compared to the asteroid and thus the center of mass of the goat and the asteroid can be approximated by the center of mass of the asteroid. The period of such an orbit would be defined by the semi-major axis:
$$
T = 2 \pi \sqrt{\frac{r^3}{\mu}},
$$
where $T$ is the orbital period, $r$ the semi-major axis and $\mu$ the gravitational parameter of the celestial body.
The maximum semi-major axis can be achieved by the goat when he would jump from the south pole parallel to the ground into an elliptical orbit with an maximum apoapsis above the north pole ($\pi r$). For this I will also assume that the goat also has a negligible size compared to $a$. Thus the maximum orbital period would be:
$$
T_j = 2 \pi \sqrt{\frac{a^3\left(1+\frac{\pi}{2}\right)^3}{\mu}}.
$$
The orbital period of a circular orbit just above the surface would be:
$$
T_s = 2 \pi \sqrt{\frac{a^3}{\mu}},
$$
so the previous period divided by this period gives,
$$
\frac{T_j}{T_s} = \left(1+\frac{\pi}{2}\right)^{3/2} \approx 4.12193848404.
$$
To check whether my first assumption is correct I will also try to find the longest solution in which the tether is put under tension. For this I will assume that the tether will always be under tension, since then I would not be able to apply symmetry around the north pole and would a part from the trajectory be a section of an orbit again. An other assumption I will make is that the tension will induce a negligible amount of torque on the asteroid, such that it will not induce a rotation. The best take off location will still be the south pole, since this will allow for the maximal path length. Assuming that the tether can not be stretched, then the tether can not do any work, since the velocity will always have to be perpendicular to it. Therefore the velocity will only be affected by gravity and by applying energy conservation the following expression for velocity can be found as a function of the radius,
$$
v = \sqrt{2\epsilon + 2\frac{\mu}{r}}
$$
To get the longest solution, then the velocity should be such that the tension should be minimal at the highest point, namely zero. This implies that the acceleration of gravity would be equal to the centripetal acceleration at the point for a circular motion with the radius equal to the length of the tether. By using this velocity to find the specific energy of the goat, then the velocity as a function of the radius becomes,
$$
v = \sqrt{\mu\left(\frac{2}{r} - \frac{2+\pi}{a(1+\pi)^2}\right)}
$$
The trajectory which is limited by the tether will be symmetric around the north pole, so only half will have to be calculated. One half will have to be slip up into two section, namely the part where the tether is partially wrapped around the asteroid and the part where the tether only pivots around the attachment point. The time it takes can be found with,
$$
dt = dl \frac{1}{dl/dt} = \sqrt{\left(\frac{\partial x}{\partial \phi}\right)^2 + \left(\frac{\partial y}{\partial \phi}\right)^2}\frac{d\phi}{v},
$$
where $x$ and $y$ are the Cartesian coordinates of the position relative to the center of mass of the asteroid and $\phi$ a temporary variable.
For the first section I will use the following temporary variable, $\alpha$, which illustrated in the image bellow,

$$
x_1 = a \left(\sin{\alpha} + (\pi - \alpha) \cos{\alpha}\right)
$$
$$
y_1 = a \left(\cos{\alpha} + (\alpha - \pi) \sin{\alpha}\right)
$$
Where $\alpha$ goes from $0$ to $\pi$.
So the time it would take to travel this first section, lets call it $t_1$, which goes from $\alpha=0$ to $\alpha=\pi$, can be calculated as follows,
$$
t_1 = \int^\pi_0{\sqrt{\left(\frac{\partial x_1}{\partial \alpha}\right)^2 + \left(\frac{\partial y_1}{\partial \alpha}\right)^2}\frac{d\alpha}{v}} = (1+\pi)\sqrt{\frac{a^3}{\mu}} \int^\pi_0{\alpha \sqrt{\frac{\sqrt{1+\alpha^2}}{2(1+\pi)^2-(2+\pi)\sqrt{1+\alpha^2}}} d\alpha} \approx 6.705301770839 \sqrt{\frac{a^3}{\mu}}.
$$
You can use wolframalpha so get to analytical solution of this integral. Due to symmetry this time will have to be accounted for twice, so the total relative time contribution of trajectory, in which the tether is partially wrapped around the asteroid, will be equal to,
$$
\frac{2t_1}{T_s} \approx 2.134363843504
$$
The second section would be a semi-circle with radius $\pi a$ and center at $(0, a)$, thus
$$
x_2 = \pi a \cos{\beta},
$$
$$
y_2 = a + \pi a \sin{\beta},
$$
where $\beta$ is the temporary variable in this case. So the time it would take to travel this second section, lets call it $t_2$, which goes from $\beta=0$ to $\beta=\frac{\pi}{2}$, can be calculated as follows,
$$
t_2 = \int^\frac{\pi}{2}_0{\sqrt{\left(\frac{\partial x_2}{\partial \beta}\right)^2 + \left(\frac{\partial y_2}{\partial \beta}\right)^2}\frac{d\beta}{v}} = \pi(1+\pi)\sqrt{\frac{a^3}{\mu}}
\int^\frac{\pi}{2}_0 {\sqrt{\frac{\sqrt{1+\pi^2+2\pi\sin{\beta}}}{2(1+\pi)^2-(2+\pi)\sqrt{1+\pi^2+2\pi\sin{\beta}}}}d\beta} \approx 10.560770017283 \sqrt{\frac{a^3}{\mu}}.
$$
This integral does not seem to have an analytical solution, so it would had to be approximated numerically. Again due to symmetry this time will have to be accounted for twice, so the total relative time contribution of trajectory, in which the tether is pivoting around the attachment point, will be equal to,
$$
\frac{2t_2}{T_s} \approx 3.361597502215
$$
So the total relative duration has a value of $\approx 5.495961345718$. So this trajectory does has a longer duration. This has probably something to do with fact that this trajectory spends more time at higher radii, such that the higher velocity does not translate into a high angular velocity, and that trajectory also has a long path length.
I also checked if there is a trajectory which would pass through the highest point $\left(0,a(1+\pi)\right)$ during which the tether is not always put under tension, but no such trajectory seemed to exist. This involved quite a few calculations, so I will not show it here. I also checked whether the tension (total acceleration minus gravity) would remain positive along the trajectory, and appears to be the case ass well. And if I would decrease the velocity by just a small amount, then the tension would become negative near the highest point. So I think that I would be able to say that this solution would be the longest jump the goat will be able to make and can be executed by jumping from the south pole vertically upwards with a velocity equal to $\sqrt{\frac{\mu}{a}}\frac{\sqrt{\pi(2\pi+3)}}{1+\pi}$.
